Download l i l i l Electrical Energy, Potential d C i and Capacitance

Electrical
l
i l Energy, Potential
i l
and
d Capacitance
C
i
Physics 1
Electric Fields and WORK
In order to bring two like charges near each other work must be
done. In order to separate two opposite charges, work must be
done Remember that whenever work gets done
done.
done, energy
changes form.
Ass tthe
e monkey
o ey does work
o o
on tthe
e pos
positive
t e ccharge,
a ge, he
e increases
c eases tthe
ee
energy
e gy o
of
that charge. The closer he brings it, the more electrical potential energy it
has. When he releases the charge, work gets done on the charge which
changes
g its energy
gy from electrical p
potential energy
gy to kinetic energy.
gy Everyy
time he brings the charge back, he does work on the charge. If he brought
the charge closer to the other object, it would have more electrical potential
energy. If he brought 2 or 3 charges instead of one, then he would have had
to do more work so he would have created more electrical potential
energy. Electrical potential energy could be measured in Joules just like any
other form of energy.
Electric Fields and WORK
Consider a negative charge moving
in between 2 oppositely charged
parallel plates initial KE=0 Final
KE= 0, therefore in this case
Work = ∆PE
We call this ELECTRICAL potential
energy, UE, and it is equal to the
amount of work done by the
ELECTRIC FORCE, caused by the
ELECTRIC FIELD over distance, d,
which
hi h iin thi
this case iis th
the plate
l t
separation distance.
Is there a symbolic relationship with the FORMULA for gravitational
potential energy?
Electric Potential
U g = mgh
g
U g → U E (or W )
m→q
g→E
h→x→d
U E (W ) = qEd
W
= Ed
q
Here we see the equation for gravitational
potential energy.
Instead of gravitational potential energy we are
talking about ELECTRIC POTENTIAL ENERGY
A charge will be in the field instead of a mass
The field will be an ELECTRIC FIELD instead of
a gravitational field
The displacement
p
is the same in any
y reference
frame and use various symbols
Putting
g it all together!
g
Question: What does the LEFT side of the equation
mean in words? The amount of Energy per charge!
Energy
gy per
p charge
g
The amount of energy per charge has a specific
name and it is called,, VOLTAGE or ELECTRIC
POTENTIAL (difference). Why the “difference”?
1 mv 2
W ∆K
∆V =
=
= 2
q
q
q
Understandingg “Difference”
Let’s say we have a proton placed
between a set of charged plates. If
the proton is held fixed at the
positive plate
plate, the ELECTRIC
FIELD will apply a FORCE on the
proton (charge). Since like charges
p the p
proton is considered to
repel,
have a high potential (voltage)
similar to being above the ground.
It moves towards the negative plate
or low potential (voltage)
(voltage). The
plates are charged using a battery
source where one side is positive
and the other is negative. The
positive side is at 9V
9V, for example
example,
and the negative side is at 0V. So
basically the charge travels through
a “change in voltage” much like a
f lli mass experiences
falling
i
a ““change
h
in height. (Note: The electron
does the opposite)
BEWARE!!!!!!
W is Electric Potential Energy (Joules)
i nott
is
V is Electric Potential (Joules/Coulomb)
a k a Voltage
a.k.a
Voltage, Potential Difference
The “other side” of that equation?
q
U g = mgh
U g → U E (or W )
m→q
g→E
h→x→d
U E (W ) = qEd
Ed
W
= Ed
q
Since the amount of energy per charge is
called Electric Potential, or Voltage, the
product of the electric field and
displacement is also VOLTAGE
This makes
Thi
k sense as it is
i applied
li d usually
ll
to a set of PARALLEL PLATES.
∆V=Ed
∆V
E
d
Example
p
A pair of oppositely charged, parallel plates are separated by
5.33 mm. A potential difference of 600 V exists between the
plates. (a) What is the magnitude of the electric field strength
between the plates? (b) What is the magnitude of the force
on an electron between the plates?
p
d = 0.00533m
∆V = 600V
E =?
qe − = 1.6 x10 −19 C
∆V = Ed
600 = E (0.0053)
E = 113,207.55 N/C
Fe
Fe
E=
=
q 1.6 x10 −19 C
Fe = 1.81x10-14 N
Example
p
Calculate the speed of a proton that is accelerated
from rest through a potential difference of 120 V
q p + = 1.6 x10 −19 C
m p + = 1.67 x10 − 27 kg
V = 120V
v=?
W ∆K
∆V =
=
=
q
q
1
2mv 2
q
2q∆V
2(1.6 x10 −19 )(120) 1.52x105 m/s
v=
=
=
−27
m
1.67 x10
Electric Potential of a Point Charge
g
Up to this point we have focused our attention solely to
that of a set of parallel plates. But those are not the
ONLY thing that has an electric field. Remember,
point charges have an electric field that surrounds
them.
them
So imagine placing a TEST
CHARGE out way from the
point charge. Will it experience
a change in electric potential
energy? YES!
Thus is also must experience a
change
g in electric p
potential as
well.
Electric Potential
Let’s use our “plate” analogy. Suppose we had a set of parallel plates
symbolic of being “above the ground” which has potential difference of
50V and a CONSTANT Electric Field.
+++++++++++
∆V = ? From 1 to 2
1
25 V
∆V = ? From 2 to 3
d
E
2
3
0V
0.5d, V= 25 V
∆V = ? From 3 to 4
12.5 V
4
0.25d, V= 12.5 V
∆V = ? From 1 to 4
37 5 V
37.5
---------------Notice that the “ELECTRIC POTENTIAL” (Voltage) DOES NOT change from 2
t 3.
to
3 They
Th are symbolically
b li ll att th
the same h
height
i ht and
d th
thus att th
the same voltage.
lt
The line they are on is called an EQUIPOTENTIAL LINE. What do you notice
about the orientation between the electric field lines and the equipotential
lines?
Equipotential
q p
Lines
So let’s say you had a positive
charge The electric field lines
charge.
move AWAY from the charge.
The equipotential lines are
perpendicular to the electric
field lines and thus make
concentric circles around the
charge. As you move AWAY
from a positive charge the
potential decreases. So
S
V1>V2>V3.
Now that we
No
e have
ha e the direction or
visual aspect of the
equipotential line understood
question is how can we
the q
determine the potential at a
certain distance away from the
charge?
r
V(r) = ?
Electric Potential of a Point Charge
g
Why the “sum” sign?
W FE x
∆V =
=
; x=r
q
q
FE r
Qq
∆V =
; FE = k 2
q
r
Qqr
Q
∆V = k 2 → k ∑
qr
r
Voltage, unlike Electric Field,
is NOT a vector! So if you
have MORE than one
charge you don’t need to use
vectors Simply add up all
vectors.
the voltages that each
charge contributes since
voltage
g is a SCALAR.
WARNING! You must use
g of the charge
g in
the “sign”
this case.
Potential of a p
point charge
g
Suppose we had 4 charges
each at the corners of a
square with sides equal to d.
d 2
If I wanted to find the potential
at the CENTER I would SUM
up all of the individual
potentials.
Thus the distance from
a corner to the center
Vcenter = k ∑
will equal :
Vcenter
r=
d 2
2
Vcenter
Q
r
q 2q − 3q 5q
= k∑( +
+
+ )
r r
r
r
5q
10q
5 2
= k( ) → k(
)→k
r
d
d 2
Electric field at the center? ( Not so easy)
If they had asked us to find the
electric field,, we first would
have to figure out the visual
direction, use vectors to break
individual electric fields into
components and use the
Pythagorean Theorem to find
the resultant and inverse
tangent to find the angle
Eresultant
Example
p
An electric dipole consists of two charges q1 = +12nC and q2
= -12nC,
12nC, placed 10 cm apart as shown in the figure.
Compute the potential at points a,b, and c.
q1 q2
Va = k ∑ ( + )
ra ra
−9
−9
12
x
10
−
12
x
10
Va = 8.99 x109 (
+
)
0.04
0.06
Va = -899 V
Example
p cont’
Vb = k ∑ (
q1 q2
+ )
rb rb
12 x10 −9 − 12 x10 −9
+
)
Vb = 8.99 x10 (
0.04
0.14
Vb = 1926.4 V
9
Vc =
0V
Since direction isn’t important, the
electric potential at “c” is zero. The
electric field however is NOT
NOT. The
electric field would point to the right.
Applications
pp
of Electric Potential
Is there any way we can use a set of plates with an electric
field? YES! We can make what is called a Parallel Plate
Capacitor and Store Charges between the plates!
Storing
St
i ChargesCh
C
Capacitors
it
A capacitor consists of 2 conductors
of any shape placed near one another
without touching
touching. It is common; to fill
up the region between these 2
conductors with an insulating material
called a dielectric
dielectric. We charge these
plates with opposing charges to
set up an electric field.
Capacitors
p
in Kodak Cameras
Capacitors can be easily purchased at a
local Radio Shack and are commonly
f
found
d in
i di
disposable
bl K
Kodak
d kC
Cameras.
When a voltage is applied to an empty
capacitor, current flows through the
capacitor and each side of the capacitor
becomes charged. The two sides have
equal and opposite charges. When the
capacitor is fully charged
charged, the current
stops flowing. The collected charge is
then ready to be discharged and when
you press the flash it discharges very
quickly released it in the form of light.
Cylindrical Capacitor
Capacitance
p
In the picture below, the capacitor is symbolized by a set of parallel
lines. Once it's charged, the capacitor has the same voltage as
the battery (1.5 volts on the battery means 1.5 volts on the
capacitor) The difference between a capacitor and a battery is
that a capacitor can dump its entire charge in a tiny fraction of a
second, where a battery would take minutes to completely
discharge itself. That's why the electronic flash on a camera uses
a capacitor -- the battery charges up the flash's capacitor over
several seconds, and then the capacitor dumps the full charge
into the flash tube almost instantly
Measuringg Capacitance
p
Let’s go back to thinking about plates!
∆V = Ed ,
∆Vα E , if d = constant
The unit for capacitance is the FARAD, F.
Eα Q Therefore
Qα ∆V
C = contant of proportionality
C = Capacitanc
C
it e
Q = CV
Q
C=
V
Capacitor
p
Geometry
y
The capacitance of a
capacitor depends on
HOW you make it.
1
Cα A Cα
d
A = area of plate
d = distance beteween plates
A
Cα
d
ε o = constant of proportionality
ε o = vacuum permittivity constant
ε o = 8.85 x10
C=
εo A
d
−12
C2
Nm 2
Capacitor
p
Problems
What is the AREA of a 1F capacitor that has a plate
separation of 1 mm?
A
C = εo
D
1 = 8.85 x10
A=
Is this a practical capacitor to build?
NO! – How can you build this then?
−12
A
0.001
1.13x108 m2
Sides =
10629 m
The answer lies in REDUCING the
AREA. But you must have a
CAPACITANCE of 1 F. How can
you keep the capacitance at 1 F
and reduce the Area at the same
ti ?
time?
Add a DIELECTRIC!!!
Dielectric
Remember, the dielectric is an insulating material placed
between the conductors to help store the charge. In the
previous example we assumed there was NO dielectric and
thus a vacuum between the plates.
A
C = kε o
d
k = Dielectric
All insulating materials have a dielectric
constant associated with it. Here now
you can reduce the AREA and use a
LARGE dielectric to establish the
capacitance at 1 F.
Usingg MORE than 1 capacitor
p
Let’s say you decide that 1
capacitor
p
will not be
enough to build what
you need to build. You
may need to use more
than 1. There are 2
basic ways to assemble
them together
„ Series – One after
a ot e
another
„ Parallel – between a set
of junctions and parallel
to each other
other.
Capacitors
p
in Series
Capacitors in series each charge each other by INDUCTION. So
they each have the SAME charge. The electric potential on the
other
th hand
h d iis di
divided
id d up amongstt th
them. IIn other
th words,
d th
the sum
of the individual voltages will equal the total voltage of the battery
or power source.
Capacitors
p
in Parallel
In a parallel configuration, the voltage is the same
because ALL THREE capacitors touch BOTH ends
of the battery. As a result, they split up the charge
amongst
g them.
Capacitors
p
“STORE” energy
gy
Anytime you have a situation where energy is “STORED” it is called
POTENTIAL. In this case we have capacitor potential energy, Uc
Suppose we plot a V vs. Q graph.
If we wanted to find the AREA we
would MULTIPLY the 2 variables
according to the equation for Area.
A = bh
When we do this we get Area =
VQ
Let’s do a unit check!
Voltage = Joules/Coulomb
Charge = Coulombs
Area = ENERGY
Potential Energy
gy of a Capacitor
p
Since the AREA under the line is a
triangle,
g , the ENERGY(area)
(
) =1/2VQ
Q
1
U C = VQ C =
2
V
This energy or area is referred
as the potential energy stored
inside a capacitor.
U C = 1 V (VC ) → 1 CV 2
2
2
2
Q
Q
U C = 1 ( )Q →
2 C
2C
Note: The slope of the line is
the inverse of the capacitance.
most common form